Why Ito Process Volatility Differs from Price Standard Deviation
Summary
The explanation distinguishes the volatility parameter in a geometric Brownian motion model from the standard deviation of the asset price itself. For an Ito process with proportional diffusion, the price is lognormally distributed, so its variance is shaped by both the diffusion and the distribution’s changing price level. The model parameter therefore should not be interpreted as the square root of the price variance at a given time.
Instead, the parameter squared gives the variance rate of log returns, with accumulated log-return variance proportional to elapsed time. This clarifies the common shorthand that volatility is standard deviation: the relevant standard deviation is typically that of returns over a specified horizon, rather than of the raw price. The discussion is limited to the stated lognormal process and does not address empirical return distributions, changing volatility, or estimation from market data.
Key ideas
- In geometric Brownian motion, the asset price follows a lognormal distribution.
- The model volatility parameter is not the standard deviation of the price level.
- The parameter squared is the variance rate of log returns in the stated model.
- Return volatility depends on the time horizon, and the explanation assumes constant volatility.
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Full text
# Why is the volatility of an Ito process not the square root of its variance? # Why is the volatility of an Ito process not the square root of its variance? The volatility $\sigma$ of an Ito process $dS_t = r S_t dt + \sigma S_t dW_t$ is not the square root of its variance. But you often hear that "volatility = standard deviation". What's going on here? ## Answer by starovoitovs (score 3) https://quant.stackexchange.com/a/46116 $S_t$ is log-normal, so indeed its variance will be different. $\sigma ^2$ is, however, the variance of the returns $\log S_t$ per unit of time since $$\log S_t \sim N\left(\log S_0 + \left(r-\frac 12 \sigma ^ 2\right) t, \sigma ^2 t\right)$$
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